Absolutly genius! When I first read about the new proof I was trying to underatand what makes it so especial since the Pythagorean theorem is surely the one with the most demostrations of all times. I really liked the new approach, and the animation as well!
This is just a preexisting geometric series that they found and they put a triangle on top. I know because I found an older version of the exact same series with completed calculations. This is more the product of dumb luck than any real intellectual insight. Look up: "math.stackexchange Is this series representation of the hypotenuse symmetric with respect to the sides of a right triangle?"
This is a very nice video. Mind Your Decisions had a very similar video out recently, but I think yours was first. That one had a good explanation of why okay to use trig in the proof. The music choice was unfortunate.
Thanks. Yes, I saw the Mind Your Decisions video that was posted after mine. He did a good job with it. Sorry you didn't like my music choice. I love how that rockin' tune enhances the coolness of the math, but I know not everyone does. (I assume you meant that *my* music choice was unfortunate, not that Presh Talwakar's choice *not* to use music was unfortunate. :-).
Very nice and satisfying. Very clear as well. Question: you mentioned the proof is special because it's trigonometric rather than geometric, and then later mention there are other trigonometric proofs. Could you clarify this point?
Hi, and thanks for the compliments on my video. Since trigonometric proofs were recently thought to be impossible without circular reasoning, it seems to me that any such proof has a special place in the storied history of the Pythagorean Theorem. In the description, I link to a 2009 paper containing another trigonometric proof which may have been the first to debunk the claim that it couldn't be done.
@@MathyJaphy Thanks for sharing the paper. I will have to read it carefully because it claims that sin^2 (x) + cos^2(x) = 1 is not equivalent to the Pythagorean theorem which I think it is: Assuming the Pythagorean theorem consider a right triangle with one of its angles equal to x and hypotenuse equal to 1. For the other direction, consider a right triangle with sides a, b, and c (c being the hypotenuse), and let x be the angle between the sides a and c. Then, by assuming the trigonometric equality, you get b^2/c^2 + a^2/c^2 = 1, which is the pythagorean theorem... So, maybe i am being silly and there is something i am not understanding... In any case, thanks for sharing that paper!
Nice video! I guess the proof does use a bit of trig, after all (sort of, by letting sin(2α) be the name of the ratio of h/c)! But it’s not necessary, as one can show that in addition to X = x₁/(1 - r) = (2ac/b)/(1 - r) you can calculate the area of the enclosing triangle in the same way: A = ab (1 + r)/(1 - r) and since we also know that A = cX/2 (ac²/b)(1 - r) = ab (1 + r)/(1 - r) c² = b² (1 + a²/b²) c² = a² + b²
Yes! Using the area formulas makes for a cleaner proof. One could also argue that trigonometry isn't needed even with the proof as presented here. It's all just ratios of side lengths of similar triangles which we happen to call "sin". This caused me to question what it even means to be a "trigonometric" proof. Even Jason Zimba's proof using the double-angle formula could be questioned (see video description for links if interested). After all, the double-angle formula comes from comparing analogous side lengths in an arrangement of similar triangles!
Can you please advise me how I can make videos like this with your animations and visuals. 1. What classes do I need to take (video editing, etc)? 2. What software did you use?
Thanks for asking! The descriptions of all my videos give credit and links to the software I use. The text animation is done in Apple Keynote. The graphical animations are done with Desmos Graphing Calculator. The resulting video clips are put together with Apple iMovie. It's not trivial, but at least all the software is free and popular, so there are online tutorials that can help you learn how to use each package individually. Making the pieces work together to create a video presentation is where most of the creativity and ingenuity is required. Yes, take a class that teaches how to use video editing software, and any math class that introduces you to Desmos.
@@MathyJaphy Thank you so much for this. I'm a professional tutor and still training to be a college professor. I would like to utilize technology like this (visual) to teach advanced students better
Excellent question! This is Euclid's Proposition 48, so yes, there is a proof by Euclid himself. You can google "Euclid Proposition 48" and find many videos showing that proof. Sadly, it relies on Proposition 47, which is the Pythagorean Theorem itself. Do you think we could reverse engineer one of the hundreds of proofs of Pythagoras to come up with a proof that doesn't rely on it? Hmmm.....
Hyperbollic geometry, but there is a curve ball here. f:=(x,y)-> , normalized R^2 vector. f(x^2-y^2, 2*x*y)=, f(x,y) forms an angle with , this doubles that angle. f(x^2-(i*y)^2, 2*x*i*y)=, inserting an i*y instead of y. My equations and whatever that thing is are very close, but not the same. There is an extra 'a' and the 'y' is negative. More than likely an artifact of presentation and my simplification.
Their proof is just a preexisting geometric series that they found and they put a triangle on top. I know because I found an older version of the exact same series with completed calculations. This is more the product of dumb luck than any real intellectual insight. Look up: "math.stackexchange Is this series representation of the hypotenuse symmetric with respect to the sides of a right triangle?"
You can't call it a proof since you're using trigonometry which a direct result of the Pythagorean theorem basically trig does not exist without the Pythagorean theorem.
Well, you could argue that it's not a trigonometric proof at all, since it only uses the sine function to refer to the ratio between the opposite leg and the hypotenuse. But if you're going to claim that it's not a proof, you'll have to point out where it uses circular reasoning. Also, I would refer you to the link in the description to Jason Zimba's paper which is a bona fide trigonometric proof that does not use circular reasoning.
@@MathyJaphy I like Zimba's proof since it does not depend upon the geometric series, which is a limiting process. Zimba's proof depends upon the sine and cosine difference angle formulae, which do not depend on Pythagoras Theorem.
Thank you! I had fun making those animations. And you're right, my derivation of X/Y is complicated. I like how the complexity dissolves away by the end, leaving the Pythagorean simplicity. However, a good proof remains straightforward throughout. I think your Variation #1 is the winner. Much prettier than mine! By the way, thanks for the mention in your latest Variation video. I am honored!
@@MathyJaphy Overall, I think that Jason Zimba's proof is the best trig proof. It relies on the difference angle formulae. It does not rely on a limiting process.
c=a+b c^2 = (a+b)^2 = [a^2 + b^2] + [2ab] (binomial expansion) c^2 a^2+b^2 The "proof" in the video is only valid in the imagination. (Pythagoras was also confused).
@@MathyJaphy The Pythagorean calculation does not include the area even if I assign c' = a + b. The point is that the sides of the triangle are affine without the area; one leg can be on earth, one inthe middle of the 'andromeda cluster, and one at the bottom of the ocean. The equaation c^2 = a^2 + b^2 obtains only if one omits the product ab (i.e., multiplication, where the area of the triangle is A = 2ab. The can only be obtained by using imaginary number where the product +/- iab is eliminatd by complex conjugation psipsi*
@@MathyJaphy That is, Fermat's Last Theorem is valid for the case n=2 for all positive real numbers c^2 a^2 + b^2 since in second order (I repeat, sigh. ad infinitum, ad nauseam) c= a + b c^2 = [a^2 + b^2] + [2ab] (Binomial Expansion, proved by Newton) [a^2 + b^2] (why) figure it out and you will be enlightened....
Absolutly genius! When I first read about the new proof I was trying to underatand what makes it so especial since the Pythagorean theorem is surely the one with the most demostrations of all times. I really liked the new approach, and the animation as well!
This is just a preexisting geometric series that they found and they put a triangle on top. I know because I found an older version of the exact same series with completed calculations. This is more the product of dumb luck than any real intellectual insight.
Look up: "math.stackexchange Is this series representation of the hypotenuse symmetric with respect to the sides of a right triangle?"
Pretty awesome... who would've imagine that the geometric series would make a surprise appearance!
Yeah, that’s my favorite part!
These type of proofs make you love maths. Awesome video!
That’s the effect I’m hoping for. Thank you!
math is circular logic, not at all real
Z’s taking high school algebra this year. He just astounded and amazed his teacher with this video :-)
Mind blowing approach man,
This is a very nice video. Mind Your Decisions had a very similar video out recently, but I think yours was first. That one had a good explanation of why okay to use trig in the proof. The music choice was unfortunate.
Thanks. Yes, I saw the Mind Your Decisions video that was posted after mine. He did a good job with it. Sorry you didn't like my music choice. I love how that rockin' tune enhances the coolness of the math, but I know not everyone does. (I assume you meant that *my* music choice was unfortunate, not that Presh Talwakar's choice *not* to use music was unfortunate. :-).
Very good
Nice💯💯💯
Very nice and satisfying. Very clear as well. Question: you mentioned the proof is special because it's trigonometric rather than geometric, and then later mention there are other trigonometric proofs. Could you clarify this point?
Hi, and thanks for the compliments on my video. Since trigonometric proofs were recently thought to be impossible without circular reasoning, it seems to me that any such proof has a special place in the storied history of the Pythagorean Theorem. In the description, I link to a 2009 paper containing another trigonometric proof which may have been the first to debunk the claim that it couldn't be done.
@@MathyJaphy Cool, Thank you.
@@MathyJaphy Thanks for sharing the paper. I will have to read it carefully because it claims that sin^2 (x) + cos^2(x) = 1 is not equivalent to the Pythagorean theorem which I think it is: Assuming the Pythagorean theorem consider a right triangle with one of its angles equal to x and hypotenuse equal to 1. For the other direction, consider a right triangle with sides a, b, and c (c being the hypotenuse), and let x be the angle between the sides a and c. Then, by assuming the trigonometric equality, you get b^2/c^2 + a^2/c^2 = 1, which is the pythagorean theorem... So, maybe i am being silly and there is something i am not understanding... In any case, thanks for sharing that paper!
Nice video! I guess the proof does use a bit of trig, after all (sort of, by letting sin(2α) be the name of the ratio of h/c)! But it’s not necessary, as one can show that in addition to
X = x₁/(1 - r) = (2ac/b)/(1 - r)
you can calculate the area of the enclosing triangle in the same way:
A = ab (1 + r)/(1 - r)
and since we also know that
A = cX/2
(ac²/b)(1 - r) = ab (1 + r)/(1 - r)
c² = b² (1 + a²/b²)
c² = a² + b²
Yes! Using the area formulas makes for a cleaner proof. One could also argue that trigonometry isn't needed even with the proof as presented here. It's all just ratios of side lengths of similar triangles which we happen to call "sin". This caused me to question what it even means to be a "trigonometric" proof. Even Jason Zimba's proof using the double-angle formula could be questioned (see video description for links if interested). After all, the double-angle formula comes from comparing analogous side lengths in an arrangement of similar triangles!
Can you please advise me how I can make videos like this with your animations and visuals.
1. What classes do I need to take (video editing, etc)?
2. What software did you use?
Thanks for asking! The descriptions of all my videos give credit and links to the software I use. The text animation is done in Apple Keynote. The graphical animations are done with Desmos Graphing Calculator. The resulting video clips are put together with Apple iMovie. It's not trivial, but at least all the software is free and popular, so there are online tutorials that can help you learn how to use each package individually. Making the pieces work together to create a video presentation is where most of the creativity and ingenuity is required. Yes, take a class that teaches how to use video editing software, and any math class that introduces you to Desmos.
@@MathyJaphy Thank you so much for this. I'm a professional tutor and still training to be a college professor. I would like to utilize technology like this (visual) to teach advanced students better
Hi there, any new video in the works??
Yes, I have one in progress and a couple more planned!
@@MathyJaphy awesome. Can't wait :)
Is there a proof that starts from the fact that if c^2=a^2+b^2, then the angle between a and b is a right angle? So the reverse way.
Excellent question! This is Euclid's Proposition 48, so yes, there is a proof by Euclid himself. You can google "Euclid Proposition 48" and find many videos showing that proof. Sadly, it relies on Proposition 47, which is the Pythagorean Theorem itself. Do you think we could reverse engineer one of the hundreds of proofs of Pythagoras to come up with a proof that doesn't rely on it? Hmmm.....
WOW
Here algebra also used. So may not say it a trigonometric proof. But a combination of both of the two. Thanks.
That's true, and geometry is used as well!
conversely you cant have convergence without euclid
Hyperbollic geometry, but there is a curve ball here.
f:=(x,y)-> , normalized R^2 vector.
f(x^2-y^2, 2*x*y)=, f(x,y) forms an angle with , this doubles that angle.
f(x^2-(i*y)^2, 2*x*i*y)=, inserting an i*y instead of y.
My equations and whatever that thing is are very close, but not the same. There is an extra 'a' and the 'y' is negative. More than likely an artifact of presentation and my simplification.
Their proof is just a preexisting geometric series that they found and they put a triangle on top. I know because I found an older version of the exact same series with completed calculations. This is more the product of dumb luck than any real intellectual insight.
Look up: "math.stackexchange Is this series representation of the hypotenuse symmetric with respect to the sides of a right triangle?"
Mainly on trig plus a bit calculus...infinite series
Point taken. And there’s still some geometry in there too.
I still think a non-geometric proof of the Pythagorean theorem is still impossible. How do you even state it without geometry?
Agreed! My assertion that this proof is "based on trigonometry rather than geometry" was poorly worded, if that's what you're referring to.
wait aren't there two proofs? I though they each came up with pure trig proofs independently.
I don't know about that. From what I've heard, they worked together on this one proof. Apparently, they've come up with more proofs since then.
You can't call it a proof since you're using trigonometry which a direct result of the Pythagorean theorem basically trig does not exist without the Pythagorean theorem.
Well, you could argue that it's not a trigonometric proof at all, since it only uses the sine function to refer to the ratio between the opposite leg and the hypotenuse. But if you're going to claim that it's not a proof, you'll have to point out where it uses circular reasoning. Also, I would refer you to the link in the description to Jason Zimba's paper which is a bona fide trigonometric proof that does not use circular reasoning.
@@MathyJaphy I like Zimba's proof since it does not depend upon the geometric series, which is a limiting process. Zimba's proof depends upon the sine and cosine difference angle formulae, which do not depend on Pythagoras Theorem.
Your video animations are cool but your derivation of X/Y is inordinately complex.
Thank you! I had fun making those animations. And you're right, my derivation of X/Y is complicated. I like how the complexity dissolves away by the end, leaving the Pythagorean simplicity. However, a good proof remains straightforward throughout. I think your Variation #1 is the winner. Much prettier than mine!
By the way, thanks for the mention in your latest Variation video. I am honored!
@@MathyJaphy Overall, I think that Jason Zimba's proof is the best trig proof. It relies on the difference angle formulae. It does not rely on a limiting process.
c=a+b
c^2 = (a+b)^2 = [a^2 + b^2] + [2ab] (binomial expansion)
c^2 a^2+b^2
The "proof" in the video is only valid in the imagination.
(Pythagoras was also confused).
In a triangle, the sum of any two side lengths is greater than the length of the third side. So your presumption, c=a+b, is incorrect.
@@MathyJaphy The Pythagorean calculation does not include the area even if I assign c' = a + b. The point is that the sides of the triangle are affine without the area; one leg can be on earth, one inthe middle of the 'andromeda cluster, and one at the bottom of the ocean. The equaation c^2 = a^2 + b^2 obtains only if one omits the product ab (i.e., multiplication, where the area of the triangle is A = 2ab. The can only be obtained by using imaginary number where the product +/- iab is eliminatd by complex conjugation psipsi*
# = a+b
#^2 = [a^2 + b^2] +[2ab]
# = 7 = 3 + 4
#^2 = 7^2 = 49 = [25] +[24] 25
(count is preserved under multiplication)
That is,
c:= a + ib
c* = a - ib
cc* = [a^2 + b^2] + a(ib) -a(ib)
(b is imaginary)
(a+b) ^2 = [a^2 + b^2] +[2ab] (binomial expansion); Fermat's Last Theorem for the case n=2
@@MathyJaphy That is, Fermat's Last Theorem is valid for the case n=2 for all positive real numbers
c^2 a^2 + b^2
since in second order (I repeat, sigh. ad infinitum, ad nauseam)
c= a + b
c^2 = [a^2 + b^2] + [2ab] (Binomial Expansion, proved by Newton)
[a^2 + b^2] (why) figure it out and you will be enlightened....
@@MathyJaphy 2nd order equations relate areas, not lengths,